Question

A mass weighing 3232 lb is attached to a spring hanging from the ceiling and comes to rest at its equilibrium position. At time tequals=​0, an external force of Upper F (t )F(t)equals=3 cosine 5 t3cos5t lb is applied to the system. If the spring constant is 1515 ​lb/ft and the damping constant is 33 ​lb-sec/ft, find the​ steady-state solution for the system. Use g equals 32 ft divided by sec squaredg=32 ft/sec2.

Answer 1

Answer:

The steady state  solution is    $x(t) = \frac{33}{157} cos(5t) + \frac{198}{157} sin(5t)$

Step-by-step explanation:

From the the question we are told that

     The mass is $m = 32lb = \frac{32* lb}{32* \frac{ft}{s^2} } = 1 slug$

       The force applied is $F(t) = 3 cos 5t$

       The spring constant is $k = 15 lb/ft$

        The damping constant is $b = 3lb-sec/ft$

        The acceleration due to gravity is $g = 32 ft/sec^2$

Generally the model differential equation for the motion of the mass attached to a string is mathematically represented as

                $F(t) = m\frac{d^2 x}{dt} + b\frac{dx}{dt} + kx$

Substituting values

             $F(t) = 1 \frac{d^2x}{dt^2} + 3\frac{dx}{dt} + 15x$

Substituting given value for F(t)

       $3cos5t = \frac{d^2x}{dt^2} + d\frac{dx}{dt} + 15x$

Applying the method of undermined coefficient in order to obtain the steady state solution

   Let

            $x(t) = Acos(5t) + Bsin(5t)$ be the steady state solution

it means that

           $x' (t) = -2A sin (5t) + 2Bcos (5t)$

And

          $x''(t) = -4Acos (5t) - 4B sin(5t)$

Now substituting this into the equation

   

$3cos5t = [-4Acos (5t) - 4B sin(5t)] + 3[-2A sin (5t) + 2Bcos (5t)] + \\\\15[Acos(5t) + Bsin(5t)]$

$3cos 5t = -4Acos (5t) - 4B sin(5t) -6Asin(5t) + 6Bcos(5t)+\\\\ 15Acos(5t) + 15Bsin(5t)$

$3cos(5t) = -4Acos(5t) + 15Acos(5t) -4Bsin(5t) +15Bsin(5t) \\\\ -6Asin(5t) + 6Bcos(5t)$

$3cos(5t) = 11Acos(5t) + 11Bsin(5t) -6Asin(5t) + 6Bcos(5t)$

Comparing coefficients  on both sides

                   Right Hand Side         Left Hand Side          

   Cos (5t)       3                                    11A + 6B

    Sin (5t)       0                                       11B -6A

 From above we have that

          $11A + 6B = 3 ---(1)$

And    

          $11B - 6A = 0 ---(2)$

From equation one we see that

        $B = \frac{6A}{11} ---(3)$

Substituting this for B in equation 1

      $11A +6 * \frac{6A}{11} = 3$

     $11A + \frac{36A}{11} = 3$

Multiply through by 11

 $121A +36A = 33$

 $157A = 33$

     $A =\frac{33}{157}$

substituting for A in equation 3

    $B = \frac{6 * \frac{33}{157} }{11}$

     $= \frac{\frac{198}{157} }{11}$

      $= \frac{198}{157} * \frac{11}{1}$

      $B = \frac{2178}{157}$

So the steady state solution is

        $x(t) = \frac{33}{157} cos(5t) + \frac{198}{157} sin(5t)$